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3 years ago

Is the process of bananas ripening a chemical change or a physical change? explain?

1 answer:

7 0

Bananas ripening is a chemical change. First, there is a change in color of peel from green to yellow then to brown. Second, there is a change in taste. You will notice a change in taste as the bananas ripen. They become sweeter and sweeter as they begin to ripen.

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Please help:/ I can’t figure out the answer

jekas [21]

Because other people cannot see what you are sending to somebody

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3 years ago

Equation Given : Al^(3+) + Na3PO4 ==> 3Na^+ + AlPO4

Helga [31]

1 mols of Aluminium ion forms 1 mol aluminium phosphate

Molar mass of AlPO_4

27+31+16(4)
58+48
106u

Moles of AlPO_4

61µg/106
0.000061/106
5.75×10^{-7}
57.5µmol

Moles of Al3+=57.5µmol

3 0

3 years ago

What is the main purpose of the second paragraph?

Svetlanka [38]

To explain your first paragraph which includes your thesis. The second paragrph supports your first pragraph

5 0

3 years ago

What are the benefits of scientists using the metric system worldwide?select all that apply

denis-greek [22]

There are several scales to determine the weight of material. The chose of the weight scale depends upon the size of the material. Like if we have to measure the molecular or atomic weight of a molecule or atom we use gm. Like that for the bigger molecules like proteins the weight of the molecule is expressed by different scale due to its big size. Like that for big material like iron rods etc we generally use kilograms or ton. There are two types of measuring units one is C.G.S. and another is S.I. The S.I. units are more acceptable worldwide than that the C.G.S. units as the digit comes in SI units are easily handable than that of CGS.

Thus metric is used to determine a large weight of compounds which is a SI unit and handable.

4 0

3 years ago

The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M

anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0

3 years ago

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